College algebra

Question

92 passengers rode in a train from city A to City B. Tickets for regular coach seats cost $121. Tickets for sleeper cars seats cost $282. The receipt for the trip totaled $19,826. How many passengers purchased each type of ticket?

Answer 1

Let x = number of coach tickets

Then 92 - x = number of sleeper tickets

 

Our equation becomes $121x + $282(92 - x) = $19826

 

Expanding, we get $121x + $25944 - 282x = $19826

 

Combining, we get -161x = -6118; multiply both sides by -1

to get 161x = 6118; and x = 38, so 92 - x = 54

 

Proof: 38 x $121 + 54 x $282 = 19826

 

Answer 2

Let c = # coach passengers

Leg s = # sleeper passengers

 

c + s = 92                    there are 92 passengers total

121c + 282s = 19826    total money was 19826

 

from c+s=92 solve for one variable in terms of the other.

c = 92-s

 

Substitute 92-s in place of c in 2nd equation to solve for s.

 

121(92-s) + 282s = 19826        distribute 121 through parenthesis

11132 - 121s + 282s = 19826   combine like terms

161s + 11132 = 19826             subtract 11132 from both sides

161s = 8694                            divide both sides by 161

s = 54

 

54 passengers rode in sleeper cars

c = 92-s = 38

38 passengers rode coach